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RoseWind [281]
2 years ago
8

How many real solutions if any, does 2x^2-3x+8=0​

Mathematics
1 answer:
fenix001 [56]2 years ago
3 0

Answer:

The quadratic equation 2\, x^{2} - 3\, x + 8 = 0 has no real solution.

Step-by-step explanation:

Rewrite the quadratic equation 2\, x^{2} - 3\, x + 8 = 0 in the standard form a\, x^{2} + b\, x + c = 0:

2\, x^{2} + (-3)\, x + 8 = 0, for which:

  • a = 2.
  • b = (-3).
  • c = 8.

The quadratic discriminant of a\, x^{2} + b\, x + c = 0 is (b^{2} - 4\, a\, c).  The quadratic discriminant of 2\, x^{2} + (-3)\, x + 8 = 0 would be:

\begin{aligned}& b^{2} - 4\, a\, c \\ =\; & (-3)^{2} - 4 \times 2 \times 8 \\ =\; & (-55)\end{aligned}.

Since the quadratic discriminant of this equation is negative, this quadratic equation has no real solution.

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Step-by-step explanation:

2+3+8+2

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